How to compare two huge xml files(50MB+) using Java Code

I want to compare two huge xml files using java code and need to find the difference of those xml files
is there any API for that

You should find third party API

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How to Append two word documents into single using   java

How to Append two word documents into single using java
we tried this but it's not append the one word document to other
source code:public class AppendTwoWordFiles {
     public static void main(String []arg)throws IOException
          FileInputStream fi=null;
          FileOutputStream fo=null;
          try {
               System.out.println("Enter the source file name u want to append");
               BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
               File f1=new File(br.readLine().toString());
               System.out.println("Enter the Destination file name ");
               File f2=new File(br.readLine().toString());
               fi = new FileInputStream(f1);
               fo = new FileOutputStream(f2,true);
               byte b[]=new byte[2];
               while((fi.read(b))!=-1);
          fo.write(b);
System.out.println("Successfully append the file");
          } catch (FileNotFoundException e) {
               // TODO Auto-generated catch block
               e.printStackTrace();
          } catch (IOException e) {
               // TODO Auto-generated catch block
               e.printStackTrace();
          finally{
          fi.close();
          fo.close();
plz reply me quickly ,,,what can i follow

Use this code ..
and give the path of the both file like this.....
source file ---- C:/workspace/Practice/src/com/moksha/ws/test/practice.text
destination file ---- C:/workspace/City/src/com/moksha/ws/test/practice1.text
import java.io.*;
public class AppendTwoWordFiles {
     public static void main(String[] arg) throws IOException {
          FileInputStream fi = null;
          FileOutputStream fo = null;
          try {
               System.out.println("Enter the source file name u want to append");
               BufferedReader br = new BufferedReader(new InputStreamReader(
                         System.in));
               File f1 = new File(br.readLine().toString());
               System.out.println("Enter the Destination file name ");
               File f2 = new File(br.readLine().toString());
               fi = new FileInputStream(f1);
               fo = new FileOutputStream(f2, true);
               byte b[] = new byte[2];
               int len = 0;
               while ((len = fi.read(b)) > 0) {
                    fo.write(b, 0, len);
               System.out.println("Successfully append the file");
          } catch (FileNotFoundException e) {
               e.printStackTrace();
          } catch (IOException e) {
               e.printStackTrace();
          } finally {
               fi.close();
               fo.close();
}

How can we get requester's user id using java code

Hi,
How can we get requester's user id using java code?
eg: If i had logged in as xelsysadm and request a resource for user uid101 on the userid field it should display uid101 and not xelsysadm.
also,
I have a resoure "A" which on revoking should also revoke resources B and C. How can it be done. Resource A, B, and C are 3 different resource objects.
==Thanks,
doki

Ok, so there is a way, but it's not available during submission. You can use the findRequests api. From the result set, get the "Requests.Consolidated Data Value" value. In this information, you will get an xml formatted data. It provides the list of users on the request on the left side after submission. After the request is completed, this value is available. Upon completion, you could get the request information, get this value, and parse the information for user ids.
In the same adapter, use the following API:
formIntf.setObjectFormData(objInstanceKey, formHash)
The formHash is a hashtable containing the field name on your object form, and the values you wish to populate it with. You could create a textarea box on your object form and populate the userids for who the request is for.
However, this makes absolutely 0 sense. When you get a request, the list of userids are listed right there on the request.
-Kevin

How to send sms to mobile from tomcatserver using java code?

Hi,
Could you please let me know that,
How to send sms to mobile from tomcatserver using java code? Please provide the code snippet.
Thanks in advance.

Yes, but something needs to send that message. You can't just take an arbitrary computer and send an SMS, it does not have the hardware to do that.
So either you have a mobile through which you do that or more likely - you use some sort of online service to do it. Whatever choice you make will determine what code you will have to write to get it done. Nobody is going to deliver the code to you, that's your job. It is also your job to figure out what service you are going to use.

How to compare two separate PDF files with certain fields identical?

Does anyone know of a way that I could compare two separate PDF files containing the same information in certain locations? Please keep in mind that they contain different information as well. I know that it could be done with importing it to excel and running VBA to format the information to be similar and use formulas to validate information is correct, but this needs to be done very quickly with multiple files through out the day. One way to explain what type of PDF file this is, is think of excel: Column A and J contain the same information but B:I contain different information. Any help or suggestions is greatly appreciated.

This unfortunately was never answered but we found a different solution rather than comparing PDF's.

How to read the contents of XML file from my java code

All,
I created an rtf report for one of my EBS reports. Now I want to email this report to several people. Using Tim's blog I implemented the email part. I am sending emails to myself based on the USERID logic.
However I want to email to different people other then me. My email addresses are in the XML file.
From the java program which sends the email, how can I read the fields from XML file. If any one has done this, Please point me to the right examples.
Please let me know if there are any exmaples/BLOG's which explain how to do this(basically read the contents of XML file in the Java program).
Thank You,
Padma

Ike,
Do you have a sample. I am searched so much in this forum for samples. I looked on SAX Parser. I did not find any samples.
Please help me.
Thank you for your posting.
Padma.

How to dynamically configure JPA setting at runtime using java code?

Hi,
I am new to EJB 3.0 and JPA. I am trying to help my company to deploy the use of JPA in EJB3.0. Currently, i am trying out with OpenJPA in IBM Webshpere Application Server 7.0. We have four different WAS servers running for testing, system integration testing, user testing and production respectively, and in each region, the configurations for data source, JDBC username and password, schema, etc are different. So i think i would need to configure the JPA setting during runtime using Java code that determines which is the environment. And apparently i am stuck with the limited knowledge i have.
1) I understand that i could override the JDBC in the persistence xml by creating entity manager using entity manager factory. But is it possible that i do similar thing by using inejction of persistence context on entity manager to obtain a container managed entity manager?
2) Alternatively, it is possible to create multiple persistence unit in the XML and inject different PU to the entity manager, am i right? But how to inject dynamically since @PersistenceContext(unitName="xxx") only accepts constant declaration.
3) Is it possible that the JDBC username and password are read from properties file?
Thanks for your help in advance!

Hi
Thanks for the reply.
Are you saying that i have to configure the data source authentication to the backend DB2 using the JAAS-J2C? Correct me if i am mistaken. I guess i am unable to do so as the application server setup only contains one data source, and the server hosted a number of applications. Every applications has their own JDBC username and password supply to the data source (the ID supplied at the DB2 side will decide which resource can access) and thus, i have to supply the username and password at runtime. Talked to the server guy and seems that it is not feasible to have seperate datasource for each applications as considered to the volume of applications hosted inside the server.
Any suggestion?

How to compare after parsing xml file

Hi,
following code, parse the input.xml file, counts how many nodes are there and writes the node name and its value on screen.
1) i am having trouble writing only node name into another file instead of writing to screen.
2) after parsing, i like to compare each node name with another .xsd file for existence.
Please keep in mind that, input.xml is based on some other .xsd and after parsing i have comparing its tag with another .xsd
Need you help guys.
thanks
* CompareTags.java
import java.io.*;
import org.w3c.dom.*;
import org.xml.sax.*;
import javax.xml.parsers.*;
import javax.xml.transform.*;
import javax.xml.transform.dom.*;
import javax.xml.transform.stream.*;
/** This class represents short example how to parse XML file,
* get XML nodes values and its values.<br><br>
* It implements method to save XML document to XML file too
public class CompareTags {
private final static String xmlFileName = "C:/input.xml";
     int totalelements = 0;
/** Creates a new instance of ParseXMLFile */
public CompareTags() {
// parse XML file -> XML document will be build
Document doc = parseFile(xmlFileName);
// get root node of xml tree structure
Node root = doc.getDocumentElement();
// write node and its child nodes into System.out
System.out.println("Statemend of XML document...");
writeDocumentToOutput(root,0);
               System.out.println("totalelements in xyz tag " + totalelements);
          System.out.println("... end of statement");
/** Returns element value
* @param elem element (it is XML tag)
* @return Element value otherwise empty String
public final static String getElementValue( Node elem ) {
Node kid;
if( elem != null){
if (elem.hasChildNodes()){
for( kid = elem.getFirstChild(); kid != null; kid = kid.getNextSibling() ){
if( kid.getNodeType() == Node.TEXT_NODE ){
return kid.getNodeValue();
return "";
private String getIndentSpaces(int indent) {
StringBuffer buffer = new StringBuffer();
for (int i = 0; i < indent; i++) {
buffer.append(" ");
return buffer.toString();
/** Writes node and all child nodes into System.out
* @param node XML node from from XML tree wrom which will output statement start
* @param indent number of spaces used to indent output
public void writeDocumentToOutput(Node node,int indent) {
// get element name
String nodeName = node.getNodeName();
// get element value
String nodeValue = getElementValue(node);
// get attributes of element
NamedNodeMap attributes = node.getAttributes();
System.out.println(getIndentSpaces(indent) + "NodeName: " + nodeName + ", NodeValue: " + nodeValue);
for (int i = 0; i < attributes.getLength(); i++) {
Node attribute = attributes.item(i);
System.out.println(getIndentSpaces(indent + 2) + "AttributeName: " + attribute.getNodeName() + ", attributeValue: " + attribute.getNodeValue());
// write all child nodes recursively
NodeList children = node.getChildNodes();
          //int totalelements = 0;
for (int i = 0; i < children.getLength(); i++) {
Node child = children.item(i);
               //     System.out.println("child value.."+child);
if (child.getNodeType() == Node.ELEMENT_NODE) {
writeDocumentToOutput(child,indent + 2);
                         if(node.getNodeName() == "DATA"){
                         totalelements = totalelements+1;}
                    //System.out.println("totalelements in DATA tag " + totalelements);
/** Parses XML file and returns XML document.
* @param fileName XML file to parse
* @return XML document or <B>null</B> if error occured
public Document parseFile(String fileName) {
System.out.println("Parsing XML file... " + fileName);
DocumentBuilder docBuilder;
Document doc = null;
DocumentBuilderFactory docBuilderFactory = DocumentBuilderFactory.newInstance();
docBuilderFactory.setIgnoringElementContentWhitespace(true);
try {
docBuilder = docBuilderFactory.newDocumentBuilder();
catch (ParserConfigurationException e) {
System.out.println("Wrong parser configuration: " + e.getMessage());
return null;
File sourceFile = new File(fileName);
try {
doc = docBuilder.parse(sourceFile);
catch (SAXException e) {
System.out.println("Wrong XML file structure: " + e.getMessage());
return null;
catch (IOException e) {
System.out.println("Could not read source file: " + e.getMessage());
System.out.println("XML file parsed");
return doc;
/** Starts XML parsing example
* @param args the command line arguments
public static void main(String[] args) {
new CompareTags();
}

hi,
check out the following links
Check this blog to extract from XML:
/people/kamaljeet.kharbanda/blog/2005/09/16/xi-bi-integration
http://help.sap.com/saphelp_nw04/helpdata/en/fe/65d03b3f34d172e10000000a11402f/frameset.htm
Check thi link for Extract from any DB:
http://help.sap.com/saphelp_nw04s/helpdata/en/58/54f9c1562d104c9465dabd816f3f24/content.htm
regards
harikrishna N

How to SaveAs the MSword file into xml file format using Java APIS?

Hello all,
I want to convert MSword file into Xml format using java APIS or any other inbuilt APIS that i can integrate in Java code.

Sorry, that doesn't make much sense.
The XML you gave is a configuration file for txt2xml utility. It doesn't represent the output format.
Are you a user of this utility?

Generate XML file dynamically using java

Hi,
I searched on this topic on the net and also in this forum but everything leads to parsing an existing file using JAXP technology. But i need to create an xml file dynamically from the data coming from a different source. I can do it manually by writting the content to a file. But i am sure there is a better way for it. I would be obliged if some one could give me and example java class which produces a simple xml file or direct me to any resource online.
Thank you in advance.

Yes you can create the xml file dynamically. Like this,
BufferedWriter bw = new BufferedWriter(new FileWriter("xyz.xml"));
bw.write(buffer.toString()); //where buffer is StringBuffer whcih contains the your xml data
bw.close();
Remember to close the BufferedWriter, otherwise it will create the file but don't write anything.
Hope this will help u.
....yogesh

File upload using java code ....?

Hi,
I have used below code to upload files into Tomcat server .
Every thing is working fine , but when i try to upload file size more then 400kb i am getting the Exception is :::::::::*javax.servlet.ServletException: javax/servlet/jsp/SkipPageException*
Let me know if any thing require to change on the below code to upload big size files..
String contentType = request.getContentType();
//out.println("Content type is :: " +contentType);
if ((contentType != null) && (contentType.indexOf("multipart/form-data") >= 0)) {
DataInputStream in = new DataInputStream(request.getInputStream());
int formDataLength = request.getContentLength();
byte dataBytes[] = new byte[formDataLength];
          int byteRead = 0;
          int totalBytesRead = 0;
          while (totalBytesRead < formDataLength) {
          byteRead = in.read(dataBytes, totalBytesRead, formDataLength);
          totalBytesRead += byteRead;
          int lastIndex = contentType.lastIndexOf("=");
          String boundary = contentType.substring(lastIndex + 1,contentType.length());
          //out.println(boundary);
          String file = new String(dataBytes);
saveFile="C:\\Program Files\\Apache Software Foundation\\Tomcat 5.5\\webapps\\StorageFolder\\Files\\UserData\\"+saveFile;
                    int pos;
                    pos = file.indexOf("filename=\"");
                    pos = file.indexOf("\n", pos) + 1;
                    pos = file.indexOf("\n", pos) + 1;
                    pos = file.indexOf("\n", pos) + 1;
                    int boundaryLocation = file.indexOf(boundary, pos) - 4;
                    int startPos = ((file.substring(0, pos)).getBytes()).length;
                    int endPos = ((file.substring(0, boundaryLocation)).getBytes()).length;
                    FileOutputStream fileOut = new FileOutputStream(saveFile);
                    //fileOut.write(dataBytes);
                    fileOut.write(dataBytes, startPos, (endPos - startPos));
                    fileOut.flush();
                    fileOut.close();
The above code is working file for small file size only ,if the file size exceeds 400kb then it will through a
*javax.servlet.ServletException: javax/servlet/jsp/SkipPageException          *

Hi,
yeah i have included above code in Try{} catch {} block even though i am getting an exception if file size exceeds 400kb . ..
when i try to upload a file more than 400kb i am getting the below exception .
HTTP Status 500 -
type Exception report
message
description The server encountered an internal error () that prevented it from fulfilling this request.
exception
javax.servlet.ServletException: javax/servlet/jsp/SkipPageException
*     org.apache.jasper.servlet.JspServlet.service(JspServlet.java:273)*
*     javax.servlet.http.HttpServlet.service(HttpServlet.java:803)*

How to store the data read from excel using java code in Ms access database

Hi. I wrote a code to read the data from excel. May i know how can i save it to Ms access database using java. Also i have many excels. So i should not connect the database using DSN. So is there any other way to achieve this?

kramish wrote:
Im pretty sure that Access does support JDBCNo it does not. It supports ODBC.
just doing a quick Google came up with some pages:
http://blog.taragana.com/index.php/archive/access-microsoft-access-database-from-java-using-jdbc-odbc-bridge-sample-code/
http://www.javaworld.com/javaworld/javaqa/2000-09/03-qa-0922-access.html
Both articles explains how to use the jdbc-odbc bridge. I think I've seen a pure jdbc driver for access but it wasn't from Microsoft and it wasn't free.
Kaj

How to change the properties of a directory using java code

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I need to change the properties of directory( websharing).
Can i do this using java code.
Regards,
Charan

I need to change the properties of
directory( websharing).
an i do this using java code.Depends on whether the server has a Java API to do it. Most likely it hasn't.

How to Read and Generate XML file from java code.

hi guys,
how to read the xml file (Condition :we know only DTD or Shema only).
How to Generate the new xml file ?(using Shema )
And one more how directly Generate the xml from DB?
Pleas with code or any URL

Using XMLbeans you can generate Java objects from an XSD schema (perhaps DTDs aswell)
Then you can create an instance of the Document object and ask it to write itself.
This will create an XML document complient to the schema.
XMLBeans generates a "type" safe DOM where you can only ever have a structure compilent to you schema.
matfud

How can I make an XML file print �, � and �

When I write an xml file by hand I can use the European alphabet and the letters �,� and �. I can also write to an ordinary text file from my Java code and still be able to display the letters as they should.
But when writing to an xml file in my Java code the letters are converted to some weird mishmash characters.
Why is this? What can I do?
// Elin

I'm assuming you are working on a Windows computer (although you can run into this on any computer.)
If you don't tell java the kind of character encoding you want, then you get an assumed desired character encoding - in Windows, the OEM characterset which does not include the characters you are trying to write.
A way to handle this is to output the kind of file you really want - that is - with the encoding you want. I suggest you might try UTF-8, which you can get by simply instantiating your output writer with the encoding string "UTF-8".

How to compare two huge xml files(50MB+) using Java Code

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